Chapter 3 - Chemical Equations and Reaction of Stoichiometry - Exercises - Concentrations of Solutions - Percent by Mass - Page 110: 61

$V=433.86ml$

$\omega =18\%=0.18$ $\rho = 1.05\frac{g}{ml}$ $m(solute)=125\%\times 65.6g = 82g$---$V=?$ $m(solute)=\omega \times m(solution)=\omega \times \rho \times V \implies V=\frac{m(solute)}{\omega \times \rho}=\frac{82g}{0.18\times 1.05\frac{g}{ml}}=433.86ml$ Hence, $433.86ml$ of this solution contains $82g$ of $NH_4Cl$.